Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove that Every compact metric space is separable$?$

Thanks in advance!!

share|improve this question

2 Answers 2

up vote 10 down vote accepted

Hint: Consider the countable family of coverings $\mathcal U_n=\left\{B\left(x,\frac1n\right)\mid x\in X\right\}$ for all $n$, use compactness to distill a countable family of points, and show it is dense.

share|improve this answer
    
Thanks for the hint, i got it. But, i feel it difficult to find this kind of open covering. There are a lot of other open coverings ; is there any trick for how to know which one would help? –  Aang Sep 4 '12 at 6:16
    
@Avatar: All of them. –  Asaf Karagila Sep 4 '12 at 6:20
    
The easiest open set in a metric space are balls; the easiest way to ensure that all points are covered is to use all points as centers; the easiest way to ensure best "granularity" is to allow arbitrarily small radii. –  Hagen von Eitzen Sep 4 '12 at 6:22
    
@Asaf Karagila: You exhibit a countable set of coverings, whic might be confusing. The straightforward covering $\{B(x,r)|x \in X, r>0\}$ should suffice. –  Hagen von Eitzen Sep 4 '12 at 6:24
    
@Avatar: But why would you care about other open covers? All you want is a countable dense set, and Asaf’s argument gives it to you. –  Brian M. Scott Sep 4 '12 at 6:25

Recall that every infinite subset $E$ of a compact metric space $X$ has a limit point in $X$. Now define a recursive process as follows: Fix some $\delta > 0$ and pick an $x_1 \in X$. Now having defined $x_1,\ldots,x_n$, if possible choose $x_{n+1}$ such that $d(x_{n+1},x_i) \geq \delta$ for all $1 \leq i \leq n$. Now we claim that this process must terminate. For suppose it does not. Consider $$E = \{x_1,x_2,x_3,\ldots \}.$$

Now by assumption of $X$ being compact, $E$ has a limit point $a \in X$ say. But then $B_\frac{\delta}{2}(a)$ can only contain at most one point of $E$, a contradiction. It follows that $X$ can be covered by finitely many neighbourhoods of radius $\delta$ about some points $x_1,\ldots,x_n$. Now what happens if you consider all balls of radius $\frac{1}{m}$ about each $x_i$ for $1 \leq i \leq n$? This will be countable, but why will it be dense in $X$? $m$ by the way runs through all positive integers.

share|improve this answer
1  
Just a minor quibble: $B_\delta (a)$ could contain more than one point of $E$. $B_{\delta / 2} (a)$ on the other hand... –  Arthur Fischer Sep 4 '12 at 11:16
    
@ArthurFischer I always get mixed up with the radius and diameter :D –  user38268 Sep 4 '12 at 13:07
    
The $n$ depends on the $m$. We get different points for each radius $1/m$. –  Stefan Hamcke Feb 9 at 22:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.